I'm new to PHP coding and I made a mistake in the code.
I have like 400 occurrences of a method that I want to change but I don't know how.
I've heard about regular expressions, but I can't apply to this case because I dont know so much of RegExp.
I have this in my code, for example
<?php echo $lang['extension_not_allowed'] ?>
I want to change all the results with $lang for $this->lang(''), for example
<?php echo $this->lang('extension_not_allowed') ?>
There's any way to do it using Ctrl+Shift+H (Global Replace) in NetBeans?
Thanks.
You can use the following regex with replacement:
(<\?php.*?)\$lang\[([^]]*)]
And the replacement:
$1\$this->\$lang($2)
Or remove the (<\?php.*?) part if the $lang may appear on lines without <?php:
\$lang\[([^]]*)]
and replace with "\$this->\$lang($1)".
See demo
A couple of notes:
(<\?php.*?) - matches and captures the text <?php and 0 or more characters other than a newline, as few as possible (with .*?)
\$lang\[ - matches $lang[ literally (note that special regex characters must be escaped)
([^]]*) - matches and captures 0 or more characters other than a ] (we are using a character class [...])
] - a literal ].
Related
I am looking for a pattern that matches everything until the first occurrence of a specific character, say a ";" - a semicolon.
I wrote this:
/^(.*);/
But it actually matches everything (including the semicolon) until the last occurrence of a semicolon.
You need
/^[^;]*/
The [^;] is a character class, it matches everything but a semicolon.
^ (start of line anchor) is added to the beginning of the regex so only the first match on each line is captured. This may or may not be required, depending on whether possible subsequent matches are desired.
To cite the perlre manpage:
You can specify a character class, by enclosing a list of characters in [] , which will match any character from the list. If the first character after the "[" is "^", the class matches any character not in the list.
This should work in most regex dialects.
Would;
/^(.*?);/
work?
The ? is a lazy operator, so the regex grabs as little as possible before matching the ;.
/^[^;]*/
The [^;] says match anything except a semicolon. The square brackets are a set matching operator, it's essentially, match any character in this set of characters, the ^ at the start makes it an inverse match, so match anything not in this set.
None of the proposed answers did work for me. (e.g. in notepad++)
But
^.*?(?=\;)
did.
Try /[^;]*/
Google regex character classes for details.
sample text:
"this is a test sentence; to prove this regex; that is g;iven below"
If for example we have the sample text above, the regex /(.*?\;)/ will give you everything until the first occurence of semicolon (;), including the semicolon: "this is a test sentence;"
Try /[^;]*/
That's a negating character class.
This was very helpful for me as I was trying to figure out how to match all the characters in an xml tag including attributes. I was running into the "matches everything to the end" problem with:
/<simpleChoice.*>/
but was able to resolve the issue with:
/<simpleChoice[^>]*>/
after reading this post. Thanks all.
this is not a regex solution, but something simple enough for your problem description. Just split your string and get the first item from your array.
$str = "match everything until first ; blah ; blah end ";
$s = explode(";",$str,2);
print $s[0];
output
$ php test.php
match everything until first
This will match up to the first occurrence only in each string and will ignore subsequent occurrences.
/^([^;]*);*/
"/^([^\/]*)\/$/" worked for me, to get only top "folders" from an array like:
a/ <- this
a/b/
c/ <- this
c/d/
/d/e/
f/ <- this
Really kinda sad that no one has given you the correct answer....
In regex, ? makes it non greedy. By default regex will match as much as it can (greedy)
Simply add a ? and it will be non-greedy and match as little as possible!
Good luck, hope that helps.
This works for getting the content from the beginning of a line till the first word,
/^.*?([^\s]+)/gm
I faced a similar problem including all the characters until the first comma after the word entity_id. The solution that worked was this in Bigquery:
SELECT regexp_extract(line_items,r'entity_id*[^,]*')
I want to match parts of a string that start with a certain character (asterisk):
abc*DEFxyz => *DEF
abc*DE*Fxyz => *DE, *F
Tried preg_match_all('/[$\*A-Z]+/', $string, $matches); But it doesn't seem to work. I get *DE*F on the 2nd example
Change your regex to this :
\*[A-Z]+
http://regexr.com?34itc
Your regex here : [$\*A-Z]+ means a string containing * and A-Z characters, not mentioning anything about start.
Try:
^[^*]*\*
which says "from the start of the line, skip over all non-asterisk characters and stop at the first"
Extending this:
s/^[^*]*\*(.*)/
Will return the remainder of the string after the asterisk. To include the asterisk, adjust like this
s/^[^*]*(\*.*)/
Here's a great tool for checking your regex: http://gskinner.com/RegExr/
Hope this helps
How would i use regular expressions to check for characters within the following string of text:
=== logo ===
I tried to use a regex tester but could come up with the correct expression for i've tried this:
/^[=]{3}$/
I want search within a string find where the text starts with 3 equal signs.
Find a string or any other characters within the equal signs.
Find 3 more equal signs.. ending the expression.
Thanks in advance.
Try using this regex:
/===[^=]+===/
If you want to capture the text, surround it in parentheses:
/===([^=]+)===/
Here's the fiddle: http://jsfiddle.net/jufXA/
If you might have equal signs in your text (but less than 3, obviously) you should instead match everything lazily (which is a tad slower):
/===(.+?)===/
Here's the fiddle: http://jsfiddle.net/jufXA/1/
How about as simple as...
/===(.+?)===/
For example:
$test = "here's ===something special===, like ===this=one===";
preg_match_all('/===(.+?)===/', $test, $matches);
var_dump($matches[1]);
Laziness is kinda virtue here: the regex engine won't advance past the first 'closing delimiter ==='. Without ?, however, you need to use negated character classes (but then again, what about ===something=like=this===?).
I prefer:
/([=]{3})\s*(.+?)\s*\1/.
This puts the text markup (three equal signs) in the beginning and then just uses a back reference for the end. It also trims your text of spaces, which is what you probably want.
I'm building this regex with a positive look ahead in it. Basically it must select all text in the line up to last period that precedes a ":" and add a "|" to the end to delimit it. Some sample text below. I am testing this in gskinner and editpadpro which has full grep regex support apparently so if I could get the answers in that for I'd appreciate it.
The regex below works to a degree but I am unsure if it is correct. Also it falls down if the text contains brackets.
Finally I would like to add another ignore rule like the one that ignores but includes "Co." in the selection. This second ignore rule would ignore but include periods that have a single Capital letter before them. Sample text below too. Thanks for all the help.
^(?:[^|]+\|){3}(.*?)[^(?:Co)]\.(?=[^:]*?\:)
121| Ryan, T.N. |2001. |I like regex. But does it like me (2) 2: 615-631.
122| O' Toole, H.Y. |2004. |(Note on the regex). Pages 90-91 In: Ryan, A. & Toole, B.L. (Editors) Guide to the regex functionality in php. Timmy, Tommy& Stewie, Quohog. * Produced for Family Guy in Quohog.
I don't think I understand what you want to do. But this part [^(?:Co)] is definitely not correct.
With the square brackets you are creating a character class, because of the ^ it is a negated class. That means at this place you don't want to match one of those characters (?:Co), in other words it will match any other character than "?)(:Co".
Update:
I don't think its possible. How should I distinguish between L. Co. or something similar and the end of the sentence?
But I found another error in your regex. The last part (?=[^:]*?\:) should be (?=[^.]*?\:) if you want to match the last dot before the : with your expression it will match on the first dot.
See it here on Regexr
This seems to do what you want.
(.*\.)(?=[^:]*?:)
It quite simply matches all text up to the last full stop that occurs before the colon.
I'm having trouble using preg_match to find and replace a string. The string of interest is:
<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span>
I need to target and replace the date, "04/30/2011" with a different date. Can someone throw me a bone a give me the regular expression to match this pattern using preg_match in PHP? I also need it to match in such a way that it only replaces up to the first closing span and not closing span tags later in the code, e.g.:
<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span><span class="hello"></span>
I'm not versed in regex, and although I've spent the last hour trying to learn enough to make this work, I'm utterly failing. Thanks so much!
EDIT: As you can see this has gotten me exhausted. I did mean preg_replace, not preg_match.
If you're after a replacement, consider using preg_replace(), something like
preg_replace('#(\d{2})/(\d{2})/(\d{4})#', '<new date>', $string);
How about this:
$toBeFoundPattern = '/([0-9][0-9])\/([0-9][0-9])\/([0-9][0-9][0-9][0-9])/';
$toBeReplacedPattern = '$2.$1.$3';
$inString = '<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span>';
// Will convert from US date format 04/30/2011 to european format 30.04.2011
echo preg_replace( $toBeFoundPattern, $toBeReplacedPattern, $inString );
and prints
EXPIRATION DATE: 30.04.2011
Patterns always begin and end with identical so called delimiter characters. Often the character / is used.
$1 references the string, which matched the first string matched by ([0-9][0-9]), $2 references be (...) and $3 the four letters matched by the last (...).
[...] matched a single character, which is one of those listed inside the brackets. E.g. [a-z] matches all lower case letters.
To use the special meaning character / inside of a pattern, you need to escape it by \ to make it be the literal slash character.
Update: Using {..} as pointed out below is shorthand for repeated patterns.
Regex should be:
(0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])[- /.](19|20)\d\d
If you want to only match one instance, this is OK. For multiple instances, use preg_match_all instead. Taken from http://www.regular-expressions.info/regexbuddy/datemmddyyyy.html.
Edit: are you looking to just search and replace inside a PHP script or do you want to do some javascript live replacement?